You could store fibs[n-1] + fibs[n-2] in fibb[n] before returning.
int fib(int n)
{
if ( n <= 1 )
{
fibb[1]=n;
return n;
}
if(fibb[n] != 0) {
return fibb[n];
}
fibb[n] = fibs[n-1] + fibs[n-2];
return fibb[n];
//return fibs(n-1)+fibs(n-2);
}
Thanks
On Sunday, October 21, 2012 7:46:43 AM UTC-4, rahul sharma wrote:
>
> #include<stdio.h>
> int fib(int n)
> {
> if ( n <= 1 )
> return n;
> return fib(n-1) + fib(n-2);
> }
>
> How can we reduce no of computations with the above code....(iterative
> solution not allowed).
>
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